Show that if a Banach space is generated by a countable set with finite dimension then it has finite dimension

Edit At first I hadn't read the title of your question so I thought that you wanted to prove Baire's theorem and I added an answer based on that. My bad.

However, I do know how to show what you have on your title using Baire's theorem so here's a sketch:

I'll leave to you to check all the details.

The following statement is a direct consequence of Baire's theorem

Theorem. If $(X,d)$ is a complete metric space and $X=bigcup_{n geq 1} X_n$, then $mathrm{int}(overline{X_n}) neq varnothing$ for some $n$.

Now I will use the theorem to show that the algebraic dimension of a Banach space is either finite or uncountable:

Assume that $X$ is a Banach space generated by ${ xi_n : n in mathbb{Z}_{>0}}$. For each $n in mathbb{Z}_{>0}$ let $X_n$ be the space generated by ${xi_1, ldots, xi_n}$. Now it's standard to show that

• $X_n$ is closed
• $mathrm{int}(X_n)= varnothing$

Since $X$ is generated by these $X_n$'s, we have $X=bigcup_{n geq 1} X_n$. Thus, by the theorem above $mathrm{int}(X_n) neq varnothing$ for some $n$, a contradiction.